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This is a tricky question. I think you are all right and this problem, because using momentum, is valid for both elastic and inelastic problem. the tricky part is, particle A cannot penetrate particle B. If, they collide, particle A moves at -97.5m/s and particle B moves with -2m/s, do you think that is possible? So if particle A initial velocity is ...


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Find the center of mass (CM) of the two connected particles. Then, determine the distance (r) from the center of mass of the part of the spring where the third particle hit it. Then, if you could allow a collision time and force while in contact, and assume the collision to be frictionless (this is might be hard for point particles and thin spring, cause ...


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The compression pulse that propagates through the metal spheres of Newton's cradle are not ordinary sound waves. They are approximate solitons (a nonlinear wave form that balances dispersion against nonlinearity). It is this property of soliton pulses that is responsible for the observed behavior. More Details: Newton's cradle is a physical manifestation ...


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The simple answer is that not only momentum but also energy needs to be conserved. This puts constraints on the number of balls that can be activated in the cradle. Note that this does not always give a unique solution either. But it enforces that $ n $ balls to $ n $ balls is a "stable" solution.


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A simplified answer is obtained by ignoring all possible effects and dependencies other that the mass of the earth (M), its angular velocity ($w$), the mass of the impacting object($m$,) and its velocity ($v$). The rotational energy of the earth is $I w^2/2$, the kinetic energy of the object is $mv^2/2$. To stop the rotation and then make it go in ...


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The equation of motion for the ball from the time it bounces till the time it hits the ground again is $$ y = v_0t - \frac{1}{2}at^2 $$ where ground level is $y=0$, and $v_0$ is the velocity going up after adjusting for the coefficient of restitution, and $t$ is the time since the bounce. This equation will take the ball through its peak and back to the ...


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Your formula's wrong. You've got $v=\frac12 at^2$, whereas that's the formula for $y$=height. Velocity's actually $v=at$ (with $a=9.8\mbox{m/sec}^2$).


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Finding the missing equations Coordinate transforms just complicate the issue. The heart of the matter is that in n dimensions you have n degrees of freedom for the velocities of the COM of each sphere, and you only have n momentum conservation equations plus one energy conservation equation. That means you need an additional n-1 equations to solve the ...


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To complete David's answer, here is a two jet event in LEP, an e+e- collider, the ALEPH detector. The experiment is running on the Z mass. The central part of ALEPH consists of several different tracking detectors. The points where charged particles interact with the tracking detectors (hits) are shown as squares and the tracks fitted to the hits are ...


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Hadronization still doesn't break conservation of energy and momentum. So getting the total energy, or the total momentum, of the quark-antiquark pair is easy: just add up the total energy and momentum of all the reaction products. To get the individual energy or momentum of one particle, i.e. just the quark (or antiquark), we rely on the fact that they ...


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Mandelstam $t = (p_1 -p_1')^2$ corresponds to the square of the momentum transfered between the two scattering particles, in an elastic scattering process. If you look at the scattering in the centre of mass frame, like you suggest, then clearly they cannot transfer any energy between them, since that would vilolate conservation of momentum.


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Firstly, as @AnubhavGoel has mentioned, I'm assuming you have mixed up the terms "elastic" and "inelastic" here as what you have said is nonphysical. Now by definition in an elastic collision total kinetic energy is conserved. In a photon, $E_k = pc$, so the total momentum of the photon is also conserved. Secondly, I'm assuming the symbol $I$ you have used ...


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The stone when hits the glass( like wind shield, or a window pane) the impulse is small as the time of contact to the force applied is more. But for the bullet the time of contact to the force applied to the glass is small so the impulse of the bullet is large. The glass is not a solid or liquid entirely. the glass is a semi solid or amorphous solid. So it ...


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"when it collides directly" is to be interpreted as a "head-on" collision? So the balls go off along the line of the original velocity direction. The question tells you that the direction of $A$ is reversed so the direction of $B$ must be in the opposite direction ie in the original direction of $A$'s velocity otherwise momentum cannot be conserved.


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I've noticed that you have referred to energy conservation a couple of times in your question and comments. So there is something that we should make very clear. Conservation of energy in general does not mean the conservation of any given kind of energy. Potential energy can turn into kinetic energy can turn into nuclear, chemical, thermal or acoustic ...


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Without further information it is not possible to predict if the two bodies will stick together or coalesce. This depends on the nature of the materials which come into contact, and the presence of any interlocking mechanism (such as used to couple carriages of a train). Momentum is always conserved in collisions; kinetic energy is not always conserved. ...


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Energy is not conserved during an inelastic collision so you cannot find $v$ using conservation of mechanical energy concept.


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Consider two objects, 1 and 2, colliding for some short time interval $\delta t$. During $\delta t$ let's ignore all forces except the contact force that 1 exerts on 2, $\vec{F}_{12}$ and the contact force that 2 exerts on 1, $\vec{F}_{21}$. As long as the objects touch each other, both of these forces exist, and by the principle of Newton's 3rd Law we know ...


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I come to think that when a mass collides with another, both of them should always have equal velocities post-collision. To paraphrase Feynmann, no matter how beautiful you may believe your reasoning to be about this, if it doesn't agree with experiment, it is wrong. If the two objects 'stick together' after the collision (the collision is totally ...


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Often when two object collide it is often represented as an instantaneous impulse exchange. However in reality this happens continuously. Namely both objects are not completely rigid and will deform during the collision, storing energy in the elastic deformation (like a spring) and dissipating energy with any inelastic deformation. During such a collision ...


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What is conserved during a collision is not velocity, but momentum (mass times velocity). If the collision is elastic, kinetic energy is also conserved: https://en.wikipedia.org/wiki/Elastic_collision. If the collision is inleastic, only momentum will be conserved: https://en.wikipedia.org/wiki/Inelastic_collision The resulting equations (which you can ...


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To calculate the minimum energy needed for the reaction the products are assumed to be stationary, i.e. the momentums are zero. With $ \pmb p^2 = E^2 - \vec p^2 = E^2 = m^2$ follows: $$({p_1^\mu}' + {p_2^\mu}' + {p_3^\mu}' + {\bar p_4^\mu}')^2 = (E_1 + E_2 + E_3 + E_4)^2 - (\vec p_1 + \vec p_2 + \vec p_3 + \vec p_4)^2 = (E_1 + E_2 + E_3 + E_4)^2 = (m_p + ...


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but still we have momentum telling us that both blocks must rise to the same height That's not true here. In the first case where the bullet is embedded, the final velocity of the bullet and the block must be identical. Since initial momentum of the two shots were the same, then the final momentum will be the same as well. Because they are connected, ...


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The simplest approach to a problem like this would assume that the collision is elastic, and that you have some knowledge of the elastic constant. But a collision between car and human is not that. Instead, let us assume that the "elbow sized object" hits the human in the mid section, and that it doesn't simply go right through him. Then the next thing that ...



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